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3 years ago

5

Carly is a catcher on the school softball team. By the end of the season she played 45% of the 40 games the team played. In how

mny games did she not play

2 answers:

faust18 [17]3 years ago

4 0

Answer:

22

Step-by-step explanation:

kirza4 [7]3 years ago

4 0

Answer:

Carly did not play 22 games

Step-by-step explanation:

You first have to figure out how many games she went to, they way you do that is 40 x .45 for the number of games she played times the percent she played.

You get 40 x .45= 18

Now you subtract the number of games she DID play from the total numbers of games from the season 40-18.

You get 40 - 18 =22

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2 years ago

What is 4/5 times 7/12

slamgirl [31]

Answer:

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Step-by-step explanation:

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2 years ago

What is the median of this set values?

Ierofanga [76]

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3 years ago

Read 2 more answers

4x-5=2x+7 midpoint segmen

Bogdan [553]

Answer:

x = 6

Step-by-step explanation:

The way to solve this is by subtracting 2x from both sides. Now we have

2x-5=7

Then we add 5 from both sides. Now we have

2x=12

We do this because we want X by itself

Then we divide by 2 on both sides and now we have

x = 6

If you have any questions feel free to ask in the comments.

7 0

3 years ago

Please help me solve this problem ASAP

DiKsa [7]

$\bold{\huge{\blue{\underline{ Solution }}}}$

<h3><u>Given </u><u>:</u><u>-</u></h3>

<u>The </u><u>right </u><u>angled </u><u>below </u><u>is </u><u>formed </u><u>by </u><u>3</u><u> </u><u>squares </u><u>A</u><u>, </u><u> </u><u>B </u><u>and </u><u>C</u>
<u>The </u><u>area </u><u>of </u><u>square </u><u>B</u><u> </u><u>has </u><u>an </u><u>area </u><u>of </u><u>1</u><u>4</u><u>4</u><u> </u><u>inches </u><u>²</u>
<u>The </u><u>area </u><u>of </u><u>square </u><u>C </u><u>has </u><u>an </u><u>of </u><u>1</u><u>6</u><u>9</u><u> </u><u>inches </u><u>²</u>

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

<u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>area </u><u>of </u><u>square </u><u>A</u><u>? </u>

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u><u> </u></h3>

The right angled triangle is formed by 3 squares

<u>We </u><u>have</u><u>, </u>

Area of square B is 144 inches²
Area of square C is 169 inches²

<u>We </u><u>know </u><u>that</u><u>, </u>

$\bold{ Area \: of \: square = Side × Side }$

Let the side of square B be x

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

$\sf{ 144 = x × x }$

$\sf{ 144 = x² }$

$\sf{ x = √144}$

$\bold{\red{ x = 12\: inches }}$

Thus, The dimension of square B is 12 inches

<h3><u>Now, </u></h3>

Area of square C = 169 inches

Let the side of square C be y

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

$\sf{ 169 = y × y }$

$\sf{ 169 = y² }$

$\sf{ y = √169}$

$\bold{\green{ y = 13\: inches }}$

Thus, The dimension of square C is 13 inches.

<h3><u>Now, </u></h3>

It is mentioned in the question that, the right angled triangle is formed by 3 squares

The dimensions of square be is x and y

Let the dimensions of square A be z

<h3><u>Therefore</u><u>, </u><u>By </u><u>using </u><u>Pythagoras </u><u>theorem</u><u>, </u></h3>

<u>The </u><u>sum </u><u>of </u><u>squares </u><u>of </u><u>base </u><u>and </u><u>perpendicular </u><u>height </u><u>equal </u><u>to </u><u>the </u><u>square </u><u>of </u><u>hypotenuse </u>

<u>That </u><u>is</u><u>, </u>

$\bold{\pink{ (Perpendicular)² + (Base)² = (Hypotenuse)² }}$

<u>Here</u><u>, </u>

Base = x = 12 inches
Perpendicular = z
Hypotenuse = y = 13 inches

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

$\sf{ (z)² + (x)² = (y)² }$

$\sf{ (z)² + (12)² = (169)² }$

$\sf{ (z)² + 144 = 169}$

$\sf{ (z)² = 169 - 144 }$

$\sf{ (z)² = 25}$

$\bold{\blue{ z = 5 }}$

Thus, The dimensions of square A is 5 inches

<h3><u>Therefore</u><u>,</u></h3>

Area of square

$\sf{ = Side × Side }$

$\sf{ = 5 × 5 }$

$\bold{\orange{ = 25\: inches }}$

Hence, The area of square A is 25 inches.

6 0

2 years ago